# Evaluation of integral related to Gamma function

In few weeks ago, I encounter some integral. At that moment, it does not require specific value. But now I have a question on evaluation. Wolframalpha suggests the following answer. $$\int_{0}^{\infty}\frac{e^{-t}-1}{t^{a}}dt=\Gamma\left(1-a\right),\quad1<a<2.$$

Well, by considering the singularity at $t=0$, it is easy to see that the integral has value.

But I don't have any experience on gamma function with negative evaluation. How can I verify that identity?

For $t > 0$, let $f_t(s) = e^{-st}$. Then

$$e^{-t} - 1 = f_t(1) - f_t(0) = \int_0^1 f_t'(s)\,ds = -t\int_0^1 e^{-st}\,ds.$$

Substituting that into the given integral, we get

$$\int_0^{\infty} \frac{e^{-t}-1}{t^a}\,dt = -\int_0^{\infty} \int_0^1 \frac{e^{-st}}{t^{a-1}}\,ds\,dt.\tag{1}$$

Since the integrand on the right hand side of $(1)$ is positive and continuous, we can interchange the order of integration, so we get

\begin{align} \int_0^{\infty} \frac{e^{-t}-1}{t^a}\,dt &= -\int_0^1 \int_0^{\infty} t^{1-a}e^{-st}\,dt\,ds \tag{$u = st$}\\ &= -\int_0^1 s^{a-2}\int_0^{\infty} u^{1-a}e^{-u}\,du\,ds \\ &= - \int_0^1 s^{a-2}\, \Gamma(2-a)\,ds\\ &= \frac{1}{1-a}\Gamma(2-a)\\ &= \Gamma(1-a). \end{align}

• I didn't think it can be proved without using complex function. Thanks. Oct 5 '16 at 14:57


$\ds{\int_{0}^{\infty}{\expo{-t} - 1 \over t^{a}}\,\dd t = \Gamma\pars{1 - a}\,,\qquad 1 < a < 2}$.

1. $\ds{\large\textsf{First Approach}}$: \begin{align} \color{#f00}{\int_{0}^{\infty}{\expo{-t} - 1 \over t^{a}}\,\dd t} & = \int_{t\ =\ 0}^{t\ \to\ \infty}\pars{\expo{-t} - 1} \,{\dd\pars{t^{1 - a}} \over 1 - a} = -\,{1 \over 1 - a}\int_{0}^{\infty}t^{1 - a}\pars{-\expo{-t}}\,\dd t \\[5mm] & = {\Gamma\pars{2 - a} \over 1 - a} = \color{#f00}{\Gamma\pars{1 - a}} \end{align}

Note that $\ds{t^{1 - a}\pars{\expo{-t} - 1} \bbox[8px,#dfd]{vanishes\ out}}$ in the limits $\ds{t \to \infty}$ and $\ds{t \to 0^{+}}$ because $\ds{1 < a <2}$.

2. $\ds{\large\textsf{Second Approach}}$:\begin{align} \color{#f00}{\int_{0}^{\infty}{\expo{-t} - 1 \over t^{a}}\,\dd t} & = \int_{0}^{\infty}\pars{\expo{-t} - 1}\ \overbrace{% {1 \over \Gamma\pars{a}}\int_{0}^{\infty}x^{a - 1}\expo{-tx} \,\dd x}^{\ds{1 \over t^{a}}}\ \,\dd t \\[5mm] & = {1 \over \Gamma\pars{a}}\int_{0}^{\infty}x^{a - 1} \int_{0}^{\infty}\bracks{\expo{-\pars{x + 1}t} -\expo{-xt}}\,\dd t\,\dd x \\[5mm] & = {1 \over \Gamma\pars{a}}\int_{0}^{\infty}x^{a - 1} \pars{{1 \over x + 1} - {1 \over x}}\,\dd x = -\,{1 \over \Gamma\pars{a}}\int_{0}^{\infty}{x^{a - 2} \over x + 1}\,\dd x \\[5mm] & \stackrel{1/\pars{x + 1}\ \mapsto\ x}{=}\,\,\, -\,{1 \over \Gamma\pars{a}}\int_{1}^{0}x\pars{{1 \over x} - 1}^{a - 2} \,\pars{-\,{\dd x \over x^{2}}} \\[5mm] & = -\,{1 \over \Gamma\pars{a}}\int_{0}^{1}x^{1 - a}\pars{1 - x}^{a - 2}\,\dd x \quad\pars{\begin{array}{l}\mbox{The integral is the}\ Beta\ Function \\ \ds{\mrm{B}\pars{2 - a,a - 1}} \end{array}} \\[2mm] & = -\,{1 \over \Gamma\pars{a}}\, {\Gamma\pars{2 - a}\Gamma\pars{a - 1} \over \Gamma\pars{1}}\quad \pars{\begin{array}{l}\ds{\mrm{B}\pars{\mu,\nu} = {\Gamma\pars{\mu}\Gamma\pars{\nu} \over\Gamma\pars{\mu + \nu}}} \\[2mm] \ds{\Gamma}:\ Gamma\ Function \end{array}} \\[2mm] & = -\,{1 \over \pi/\braces{\Gamma\pars{1 -a}\sin\pars{\pi\bracks{1 - a}}}}\, \pi\sin\pars{\pi\bracks{a - 1}}\quad \pars{\begin{array}{l}\mbox{We used twice} \\ Euler\ Reflection \\ Formula \\[2mm] \mbox{and}\ \ds{\Gamma\pars{1} = 1}\end{array}} \\ & = \color{#f00}{\Gamma\pars{1 - a}} \end{align}

• Nice alternative proof. Thanks. Oct 6 '16 at 3:57
• @WillKwon $\pars{\bullet\,\,\,\bullet \atop {\mid \atop {\large \smile}}}$. Glad it was helpful for you. It's always nice to have different points of view. Oct 6 '16 at 4:01
• +1 for the solutions. The first would have been the way I would have pursued it. Oct 21 '16 at 21:31